A cell of internal resistance $r$ drives current through an external resistance $R$ . The power delivered by the cell to the external resistance will be maximum when:
JEE MAIN 2019, Medium
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Current $i=\frac{E}{r+R}$
Power generated in $R$
$P=i^{2} R$
$P=\frac{E^{2} R}{(r+R)^{2}}$
For maximum power $\frac{\mathrm{dP}}{\mathrm{dR}}=0$
$E^{2}\left[\frac{(r+R)^{2} \times 1-R \times 2(r+R)}{(r+R)^{4}}\right]=0$
$\Rightarrow r=R$
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